A ball is launched from ground level. Its height in meters t seconds after launch is modeled by \(\mathrm{H(t) =...

1. TRANSLATE the problem information

• Given information:
  • Height function: \(\mathrm{H(t) = -5(t - 2)^2 + 20}\)
  • Ball is on the ground at \(\mathrm{t = 0}\) and at \(\mathrm{t = s}\), where \(\mathrm{s \gt 0}\)
  • Need to find the value of s
• What "on the ground" tells us: When the ball is on the ground, its height is zero, so \(\mathrm{H(t) = 0}\)

2. TRANSLATE the key condition

• Since the ball is on the ground when \(\mathrm{H(t) = 0}\), we need to solve:
  \(\mathrm{-5(t - 2)^2 + 20 = 0}\)

3. SIMPLIFY the equation

• Subtract 20 from both sides: \(\mathrm{-5(t - 2)^2 = -20}\)
• Divide both sides by -5: \(\mathrm{(t - 2)^2 = 4}\)
• Take the square root of both sides: \(\mathrm{t - 2 = \pm 2}\)

4. CONSIDER ALL CASES from the square root

• When \(\mathrm{t - 2 = +2}\): \(\mathrm{t = 4}\)
• When \(\mathrm{t - 2 = -2}\): \(\mathrm{t = 0}\)
• So the ball is on the ground at \(\mathrm{t = 0}\) and \(\mathrm{t = 4}\)

5. APPLY CONSTRAINTS to identify the final answer

• We're told the ball hits the ground at \(\mathrm{t = 0}\) and \(\mathrm{t = s}\), where \(\mathrm{s \gt 0}\)
• Since our solutions are \(\mathrm{t = 0}\) and \(\mathrm{t = 4}\), and s must be positive: \(\mathrm{s = 4}\)

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that "on the ground" means \(\mathrm{H(t) = 0}\). Instead, they might try to work directly with the given function or misinterpret what they're solving for.

Without this crucial translation, they can't set up the correct equation and end up confused about how to proceed. This leads to confusion and guessing.

Second Most Common Error:

Poor CONSIDER ALL CASES execution: Students solve \(\mathrm{(t - 2)^2 = 4}\) but only consider the positive square root, getting \(\mathrm{t - 2 = 2}\), so \(\mathrm{t = 4}\). They miss that square roots have both positive and negative solutions.

This incomplete work means they don't verify that \(\mathrm{t = 0}\) is also a solution, potentially leading them to doubt their answer or select Choice B (2) if they misinterpret the vertex.

The Bottom Line:

This problem tests whether students can translate a physics context into mathematical language and then systematically solve a quadratic equation. The key insight is recognizing that "on the ground" creates the condition \(\mathrm{H(t) = 0}\), which transforms the problem into a straightforward algebraic exercise.